Find for which $\alpha \in \mathbb{R}$, the line $y=8x+\alpha$ is tangent to the curve $x^4+y^4=1$.
Firstly, I calculated the tangent to the curve, which is $(4x^3, 4y^3)$, and if the line is tangent to the curve, then exists $(x,y)$ such that $x^4+y^4=1$, and $y=8x+\alpha$ and $4tx^3=x, 8x+\alpha=4ty^3$ for some $t$. This direction didn't yield any results, so I tried to parametrize the curve, but am not sure that $\gamma(t)=(\sqrt{cost}, \sqrt{sint})$ is the direction, but this was problematic since this isn't well defined when $cost<0$ or $sint<0$ and using the parametrization $\gamma (t)=(\sqrt{|cost|}, \sqrt {|sint|}$ isn't one to one and onto.
What's the best direction to use?
Differentiating the curve $x^4+y^4=1$ gives slope $-\frac{u^3}{v^3}$ at $(u,v)$. So we need $v^3=-\frac{u^3}{8}$ and hence $v=-\frac{u}{2}$. The point lies on the curve so $u^4=\frac{16}{17}$. Hence $u=\pm\frac{2}{k}$, where $k=17^{1/4}$.
For the point $(u,v)=(\frac{2}{k},-\frac{1}{k})$ we have $\alpha=v-8u=-k^3$, for the point $(u,v)=(-\frac{2}{k},\frac{1}{k})$ we have $\alpha=v-8u=k^3$. So the two possible values of $\alpha$ are $\pm17^{3/4}\approx\pm8.37214$.